How do I show $(\mathbb{Z} / 12 \mathbb{Z})^{*} \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$ I need to show that $(\mathbb{Z} / 12 \mathbb{Z})^{*} \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$.
But I don't even know where to begin. Is there a smart way to show this where I don't have to come up with an actual isomorphism between the two groups? Showing that two groups are isomorphic is one of the hardest problems that I have encountered since it's often difficult to invent an isomorphism between the two groups.

I have encountered a similar exercise where I had to show
$(\mathbb{Z} / 14 \mathbb{Z})^{*}\cong(\mathbb{Z} / 2 \mathbb{Z}) \times(\mathbb{Z} / 3 \mathbb{Z})$
But this was much simpler since I showed that $|(\mathbb{Z} / 14 \mathbb{Z})^{*}|=\varphi(14)=6$ and that $(\mathbb{Z} / 14 \mathbb{Z})^{*}$ is a cyclic group.
So I had a theorem that said that a cyclic group of order 6 is isomorphic to the quotient group $\mathbb{Z} /6 \mathbb{Z}$.
Furthermore, $6=2\cdot3$ and $gcd(2,3)=1$ so the CRT tells me that $\mathbb{Z} /6 \mathbb{Z}\cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$.
That is $(\mathbb{Z} / 14 \mathbb{Z})^{*}\cong\mathbb{Z} /6 \mathbb{Z}\cong\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z}$
However, none of these methods work in the case above since $gcd(2,2)\neq1$ and the group $(\mathbb{Z} / 12 \mathbb{Z})^{*}$ is not cyclic.
 A: Do you know what $H=(\mathbb{Z}/12\mathbb{Z})^*$ looks like as a set?
You'll notice it has four elements and since it's a subgroup of an abelian group, it's abelian. That only leaves a couple of possibilities:
If $H$ has an element of order $4$ then it's isomorphic to $\mathbb{Z}/4\mathbb{Z}$, and if it doesn't it's isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ (why?). So at this point you can just check the order of the elements manually.
A: The most obvious way to show this is by using the Chinese Remainder Theorem to see that ${\mathbb Z}_{12} \cong {\mathbb Z}_4 \times {\mathbb Z}_3$ (as rings) and therefore also ${\mathbb Z}_{12}^* \cong {\mathbb Z}_4^* \times {\mathbb Z}_3^*$ (as multiplicative groups). Now ${\mathbb Z}_4^*$ and ${\mathbb Z}_3^*$ both have two elements, so both are cyclic of order $2$.
This approach generalizes to finding ${\mathbb Z}_m^*$ and reduces it to finding ${\mathbb Z}_{p^k}^*$ for a prime $p$. (Which is a cyclic group of order $(p-1)p^{k-1}$, except if $p = 2$. There ${\mathbb Z}_2^* \cong \{1\}$, ${\mathbb Z}_4^* \cong {\mathbb Z}_2$, and ${\mathbb Z}_{2^k}^* \cong {\mathbb Z}_2 \times {\mathbb Z}_{2^{k-2}}$ for $k \geq 3$).
If for some reason this is not an acceptable argument, then the answer of course depends on what acceptable arguments are.
A more primitive way of seeing that ${\mathbb Z}_{12}^* \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ is noting that the elements of ${\mathbb Z}_{12}^*$ are $\overline{1}$, $\overline{5}$, $\overline{7}$, and $\overline{11}$. Then work out the multiplication table and see that it is the same, after renaming, as that of ${\mathbb Z}_2 \times {\mathbb Z}_2$.
